Answer:
p_{f} = 6 m / s
Explanation:
We can solve this exercise using conservation of momentum. For this we define a system formed by the two balls, so that the forces during the collision have been intense and the moment is preserved
Initial instant. Before the crash
p₀ = m v +0
Final moment. Right after the crash
= (m + m) v_{f}
how the moment is preserved
p₀ = p_{f}
m v = 2 m v_{f}
v_{f} = v / 2
we calculate
v_{f} = 12/2
p_{f} = 6 m / s
"Fig is attacted with answer"
Answer:
a) d = 33.72 m
b) = 26 m/s
c) β = 71.08°
Explanation:
a)
When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.
Given data:
time = t = 4.00 s
Height = h = 20 m
Angle = θ = 60°
Horizontal distance = d = ?
Using 2nd equation of motion
-20 = (4) + 0.5(-9.8)(4)²
(4) = 58.4
= 14.6 m/s
This is vertical component of velocity when the ball is on the roof. To calculate the Final velocity and horizontal component, we use
=
/ sinθ
= 14.6 / sin 60
= 16.86 m/s
=
cosθ
= 16.86 cos 60
= 8.43 m/s
To calculate the horizontal distance
d = t
d = (8.43)(4)
d = 33.72 m
b)
We know the values of Landing angle, height of roof, time of flight. In part a, We calculate the landing velocity of the ball and also its horizontal and vertical component. As the ball followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.
So,
=
but the vertical component of velocity vary with and there is an acceleration along vertical direction which is equal to gravitation acceleration g.
So,
g = ( -
) / t
9.8 = 14.6 - ) / 4
= 24.6 m/s
=
=
= 26 m/s
c)
cos β = /
β = cos⁻¹ (8.43 / 26)
β = 71.08°
